Regressions

Simple linear regression

Y_i = \beta_0 + \beta_1 x_i + \epsilon_i

Ideal for predicting a continuous response variable from a single predictor variable: “How does y change as x changes, when the relationship is linear?”

Multiple linear regression

Y_i = \beta_0 + \beta_1 x_{1i} + \beta_2 x_{2i} + ... + \beta_k x_{ki} + \epsilon_i

“How does y change as x_1, x_2, …, x_k change?”

Fitting a nonlinear model

Linear relationships are simple to interpret since the rate of change is constant.

“As one changes, the other changes at a constant rate.”

Nonlinear relationships often involve exponential, logarithmic, or power functions.

“As one changes, the other changes at a rate that is not proportional to the change in the other.”

Nonlinear relationships: exponents

• x^2 is the square of x.
• x^3 is the cube of x.
• x^a is x raised to the power of a.

In a relationship where y is a function of x^a, as y increases, x increases at a rate that is equal to x to the power of a.

Nonlinear relationships: logarithms

• log_e(x) is the natural logarithm of x.
• log_{10}(x) is the common logarithm of x.
• log_a(x) is the logarithm of x to the base a.

Interpretation:

• If \log_a(y) = x: as x increases, y increases at a rate of y = a^x.
• If y = \log_a(x): as y increases, x also increases, at x = a^y.

Exponents and logarithms

Dealing with nonlinearity

Transformations

Often, a nonlinear relationship may be transformed into a linear relationship by applying a transformation to the response variable or the predictor variable(s).

• Logarithmic: y = \log(x)
• Exponential: y = e^x
• Square-root: y = \sqrt{x}
• Inverse: y = \frac{1}{x}

Exponential decay relationship

Response variable decreases and approaches limit as predictor variable increases.

y = a \cdot e^{-b_x}

Examples: radioactive decay, population decline, chemical reactions.

Asymptotic relationship

Response variable increases and approaches a limit as the predictor variable increases.

y = a + b(1 - e^{-cx})

Examples: population growth, enzyme kinetics.

Logistic relationship

An S-shaped relationship, where the response variable is at first exponential, then asymptotic.

y = c + \frac{d-c}{1+e^{-b(x-a)}}

Examples: growth of bacteria, disease spread, species growth.

Curvilinear relationship

Response variable changes in a variety of ways as the predictor variable changes.

y = a + bx + cx^2 + dx^3 + ...

Examples: food intake, drug dosage, exercise.

Transformations: Exponential decay

Transformations: Exponential decay

Transformations: Asymptotic relationship

Transformations: Asymptotic relationship

Transformations: Logistic relationship

Transformations: Logistic relationship

Transformations: Curvilinear relationship

Transformations: Curvilinear relationship

Did the transformations work?

• To a certain extent…
• Problems:
  • Relationships typically do not meet the linear assumption, but seem “ok” for other assumptions.
  • Poor fit to the data (over or underfitting in some areas).
  • Difficult to interpret the results.

Nonlinear regression

• A way to model complex (nonlinear) relationships.
  • i.e. phenomena that arise in the natural and physical sciences e.g. biology, chemistry, physics, engineering.
• At least one predictor is not linearly related to the response variable.

Wait!

Model

Y_i = \beta_0 + \beta_1 x_i + \beta_2 x_i^2 + ... + \beta_k x_i^k + \epsilon_i

where k is the degree of the polynomial.

• The model is still linear in the parameters \beta and can be fitted using least squares.
• Instead of multiple predictors, we have multiple terms of the same predictor.
• Can still be fit using lm().

The data

See Slide 11 for the relationship and mathematical expression.

Fitting the model (linear)

Y_i = \beta_0 + \beta_1 x_i + \epsilon_i

Fitting the model (poly order 2)

Y_i = \beta_0 + \beta_1 x_i + \beta_2 x_i^2 + \epsilon_i

Fitting the model (poly order 3)

Y_i = \beta_0 + \beta_1 x_i + \beta_2 x_i^2 + \beta_3 x_i^3 + \epsilon_i

Fitting the model (poly order 10)

Y_i = \beta_0 + \beta_1 x_i + \beta_2 x_i^2 + ... + \beta_10 x_i^{10} + \epsilon_i

Limitations

• Meaning of the coefficients is not always clear.
• Extrapolation can be dangerous.
• Extra terms can lead to overfitting and are difficult to interpret:

Call:
lm(formula = response ~ poly(predictor, 10), data = asymptotic)

Residuals:
     Min       1Q   Median       3Q      Max 
-17.1659  -8.6908  -0.0494   8.8003  16.4012 

Coefficients:
                      Estimate Std. Error t value Pr(>|t|)    
(Intercept)             79.818      1.552  51.426  < 2e-16 ***
poly(predictor, 10)1   159.368     11.084  14.378  < 2e-16 ***
poly(predictor, 10)2  -106.939     11.084  -9.648 5.37e-12 ***
poly(predictor, 10)3    48.570     11.084   4.382 8.28e-05 ***
poly(predictor, 10)4   -19.411     11.084  -1.751   0.0876 .  
poly(predictor, 10)5     1.193     11.084   0.108   0.9148    
poly(predictor, 10)6    -2.769     11.084  -0.250   0.8040    
poly(predictor, 10)7    -1.343     11.084  -0.121   0.9042    
poly(predictor, 10)8    -4.009     11.084  -0.362   0.7195    
poly(predictor, 10)9    -2.851     11.084  -0.257   0.7984    
poly(predictor, 10)10    5.769     11.084   0.520   0.6056    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 11.08 on 40 degrees of freedom
Multiple R-squared:  0.8897,    Adjusted R-squared:  0.8621 
F-statistic: 32.26 on 10 and 40 DF,  p-value: 4.846e-16

But:

• Easy to fit: just add polynomial terms to the model.
• Simple to perform: use lm().

Fitting a nonlinear model

If you have some understanding of the underlying relationship (e.g. mechanistic process) between the variables, you can fit a nonlinear model.

Assumptions

Like the linear model, the nonlinear model assumes:

• Error terms are normally distributed (Normality).
• Error terms are independent (Independence).
• Error terms have constant variance (Homoscedasticity).

Basically:

\epsilon_i \sim N(0, \sigma^2)

Estimating the model parameters

• The parameters are estimated using the method of least squares.
• For nonlinear models, a nonlinear optimization algorithm is used to find the best fit, rather than ordinary least squares, e.g.:
  • Gauss-Newton algorithm
  • Levenberg-Marquardt algorithm
• This can only be performed iteratively and depends on a “best guess” of the parameters as a start.
  • i.e. we need to provide a starting point for a nonlinear least squares algorithm to begin.

Source: Wikipedia

Implementation

use nls() function in R.

• formula: a formula object, with the response variable on the left of a ~ operator, and the predictor variable(s) on the right.
• data: a data frame containing the variables in the model.
• start: a named list of starting values for the parameters in the model.

Finding starting values

y = a + b(1 - e^{-cx})

• a is value of y when x = 0.
• b is the upper limit: the maximum value of y.
• c is the rate of change: the rate at which y approaches the upper limit.

First guess

y = a + b(1 - e^{-cx})

Check the fit

Interpretation

Formula: response ~ a + b * (1 - exp(-c * predictor))

Parameters:
   Estimate Std. Error t value Pr(>|t|)    
a -14.51755    6.64162  -2.186   0.0337 *  
b 113.03797    6.44716  17.533  < 2e-16 ***
c   0.62962    0.07142   8.816 1.32e-11 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 10.21 on 48 degrees of freedom

Number of iterations to convergence: 4 
Achieved convergence tolerance: 1.178e-06

• The model is significant since the p-value is less than 0.05 for all parameters.
• The parameterised model is:

y = -14.5 + 113.04(1 - e^{-0.63x}) The R-square value is not reported for nonlinear models as the sum of squares is not partitioned into explained and unexplained components.

Recap on logistic relationship

y = c + \frac{d-c}{1+e^{-b(x-a)}}

Recap on logistic relationship

y = c + \frac{d-c}{1+e^{-b(x-a)}}

Finding the starting values

y = c + \frac{d-c}{1+e^{-b(x-a)}}

• c is the lower limit: the minimum value of y.
• d is the upper limit: the maximum value of y.
• a is the value of x when y is halfway between the lower and upper limits.
• b is the rate of change: the rate at which y approaches the upper limit.

Finding the starting values

y = c + \frac{d-c}{1+e^{-b(x-a)}}

• c is the lower limit: the minimum value of y.
• d is the upper limit: the maximum value of y.
• a is the value of x when y is halfway between the lower and upper limits.
• b is the rate of change: the rate at which y approaches the upper limit.

Self-starting functions

• The nls() function requires a formula and starting point(s) for the parameters.
  • How about starting to nope out…

Revisiting the logistic model

y = c + \frac{d-c}{1+e^{-b(x-a)}}

• input: the predictor variable.
• Asym: the upper limit.
• xmid: the value of x when y is halfway between the lower and upper limits.
• scal: the rate of change.

The equation ia different: see ?SSlogis:

y = \frac{Asym}{1+exp \frac{xmid-input}{scal}}

What does the fit look like?

Check the fit

Interpretation

Formula: response ~ SSlogis(predictor, Asym, xmid, scal)

Parameters:
      Estimate Std. Error t value Pr(>|t|)    
Asym 310.64727    4.62579   67.16   <2e-16 ***
xmid   4.92715    0.07142   68.99   <2e-16 ***
scal   1.34877    0.05418   24.90   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 10.22 on 48 degrees of freedom

Number of iterations to convergence: 1 
Achieved convergence tolerance: 6.626e-07

Comparing manual fit to self-starting function

The self-starting function for the asymptotic model is SSasymp().

Comparing outputs:

In some cases, the fits are identical, but in others, they are not.

Summary

• When fitting a nonlinear model, there are three possible approaches:
  1. Linearize the model by transforming the response variable or predictor variable(s):
     • Fit: easy/difficult
     • Interpret: difficult
  2. Approximate the model by adding polynomial terms:
     • Fit: easy
     • Interpret: difficult
  3. Fit the model using a nonlinear least squares algorithm:
     • Fit: difficult
     • Interpret: easy

Example: polynomial regression

Prediction quality

We can use prediction quality metrics to compare the fits.

• Akaike information criterion (AIC) and Bayesian information criterion (BIC).
  • Useful for comparing model fits.
• root mean squared error (RMSE) and mean absolute error (MAE).
  • Useful for assessing the quality of the fit.

AIC and BIC

Use the broom package to extract the AIC and BIC values from the model fits.

# A tibble: 4 × 3
  model    AIC   BIC
  <chr>  <dbl> <dbl>
1 linear  453.  459.
2 poly2   409.  416.
3 poly3   392.  402.
4 poly10  402.  425.

• The smaller the AIC or BIC, the better the fit compared to other models.
• However, for better performance, cross-validation is recommended as it explains how well the model will perform on new data, rather than just assessing the fit to the data.

Cross-validation

1. Split the data into training and testing sets.
2. Fit the model to the training set.
3. Predict the response variable using the testing set.
4. Calculate the RMSE or MAE between the predicted and observed values.
5. Repeat for each fold of the data.
6. Average the RMSE or MAE across all folds.
7. The model with the lowest RMSE or MAE is the best model.

Looks like a lot of work, but it’s easy in R using the caret package.

Performing cross-validation on the polynomial fits

From the results, the cubic model has the lowest RMSE and MAE, so is the best model.